I have been working on the original problem and have narrowed it down to proving that :
$\displaystyle lim_{n \to \infty} \sum_{i=1}^{M}\left(k_{i}\cos\left(n\theta_{i}-d_{i}\right)\right)$ does not exist (or if exists, is not equal to 0) ,$\forall d_{i}$ & $\theta_{i} \in \mathbb{R}$ ,$k_{i} \in \mathbb{R^{+}}$, $M \in \mathbb{N}\geq {2}$
Given : $\theta _{i}$ are independent in the sense that $a\cos\left(n\theta_{i1}+\phi_{1}\right)+b\cos\left(n\theta_{i2}+\phi_{2}\right)$ cannot be expressed in terms of $c\cos\left(n\gamma+\phi\right)$
Intuitively it makes sense because all the cosines will oscillate and hence their linear combination is also supposed to oscillate but how do I prove this rigorously?
Some efforts : If the limit exists and equal to $L$ then $\displaystyle lim_{n \to \infty} \sum_{i=1}^{M}\left(k_{i}\cos \left(p\theta_{i}\right)\cos\left(n\theta_{i}-d_{i}\right)\right)$ is also equal to $L$ , $\forall p \in \mathbb{N}$ this can be obtained by just changing the variable $n\to n+2p$ and adding with the original limit. We can repeat the process $r$ times to obtain product of $r$ such cosines in front of each summation term. But I don't see where I am going with this.
Disclaimer : I am not well versed with Mathematics so please point out the faults, in the comments. I'll be highly obliged.
Let the limit exists and be equal to $L$
Let $q=n-2$
So, $\displaystyle \begin {equation}L=lim_{n \to \infty} \sum_{i=1}^{M}\left(k_{i}\cos\left(n\theta_{i}-d_{i}\right)\right)\tag{i}\end {equation}$ $\displaystyle \begin{equation}\Rightarrow L=lim_{n \to \infty} \sum_{i=1}^{M}\left(k_{i}\cos\left(\left(q+2\right)\theta_{i}-d_{i}\right)\right)\end{equation}$
$n\to \infty \Rightarrow n-2\to\infty\Rightarrow q\to \infty$ $\displaystyle \Rightarrow L=lim_{q \to \infty} \sum_{i=1}^{M}\left(k_{i}\cos\left(\left(q+2\right)\theta_{i}-d_{i}\right)\right)$
Replacing the dummy variable, $\displaystyle \begin {equation}L=lim_{n \to \infty} \sum_{i=1}^{M}\left(k_{i}\cos\left(\left(n+2\right)\theta_{i}-d_{i}\right)\right)\tag{ii}\end {equation}$
Adding (ii) && (i), $\displaystyle \begin {equation}2L=lim_{n \to \infty} \sum_{i=1}^{M}\left(k_{i}\left[\cos\left(\left(n+2\right)\theta_{i}-d_{i}\right)+\cos\left(n\theta_{i}-d_{i}\right)\right]\right)\end {equation}$ Using $\begin{equation}\cos a +\cos b = 2\cos\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)\end{equation}$,
$\displaystyle \begin {equation}L=lim_{n \to \infty} \sum_{i=1}^{M}\left(k_{i}\cos\theta_{i}\cos\left(\left(n+1\right)\theta_{i}-d_{i}\right)\right)\end {equation}$
Replacing $r=n+1$ and re-replacing $r=n$,
$\displaystyle \begin {equation}L=lim_{n \to \infty} \sum_{i=1}^{M}\left(k_{i}\cos\theta_{i}\cos\left(n\theta_{i}-d_{i}\right)\right)\tag{iii}\end {equation}$
Multiplying (i) by $\cos\theta_{1}$ and subtracting it from (iii),
$\displaystyle \begin {equation}L\left(1-\cos\theta_{1}\right)=lim_{n \to \infty} \sum_{i=2}^{M}\left(k_{i}\left(\cos\theta_{i}-\cos\theta_{1}\right)\cos\left(n\theta_{i}-d_{i}\right)\right)\tag{iv}\end {equation}$
Now replacing $L\left(1-\cos\theta_{1}\right)$ by $L1$ and $k_{i}\left(\cos\theta_{i}-\cos\theta_{1}\right)$ by $k_{i1}$,
$\displaystyle \begin {equation}L1=lim_{n \to \infty} \sum_{i=2}^{M}\left(k_{i1}\cos\left(n\theta_{i}-d_{i}\right)\right)\tag{v}\end {equation}$
As the algorithm to transform $\left(i\right)\to\left(v\right)$ did not assume any limitation on the values of $k_{i}$,thus same algorithm could be applied to transform $\left(v\right)\to\left(vi\right)$,
$\displaystyle \begin {equation}L2=lim_{n \to \infty} \sum_{i=3}^{M}\left(k_{i2}\cos\left(n\theta_{i}-d_{i}\right)\right)\tag{vi}\end {equation}$where $L2=L1\left(1-\cos\theta_{2}\right)$ and $k_{i2}=k_{i1}\left(\cos\theta_{i}-\cos\theta_{2}\right)$
So it follows that $L_{j}=L_{j-1}\left(1-\cos\theta_{j}\right)$ and $k_{ij}=k_{i j-1}\left(\cos\theta_{i}-\cos\theta_{j}\right)$
By repeating the transformations, we have :
$\displaystyle \begin {equation}L_{M-1}=lim_{n \to \infty} \sum_{i=M}^{M}\left(k_{i M-1}\cos\left(n\theta_{i}-d_{i}\right)\right)\tag{vii}\end {equation}$
Now, $k_{M M-1} = \prod_{i=1}^{M-1}\left(\cos\theta_{M}-\cos\theta_{i}\right) \neq 0$, as $\theta_{i}$ are independent
Let $L_{F}=\frac{L_{M-1}}{k_{M M-1}}$, Using (vii) :
$\displaystyle \begin {equation}L_{F}=lim_{n \to \infty} \cos\left(n\theta_{M}-d_{M}\right)\tag{viii}\end {equation}$
As $M\geq2$, we can always choose $\theta_{M}\neq 2p\pi$, $p\in\mathbb{N}$
But equation (viii) is incorrect (because such a limit cannot exist$^{1}$) hence by contradiction equation (i) is incorrect or such a limit does not exist.
Thus proved.
1 : I think that's the weakest point of the proof?